All categories

2 years ago

What force on a 936kg elevator that is falling freely at 9.8 m/s^2

Physics

1 answer:

nataly862011 [7]2 years ago

5 0

Answer:

9172.8 N

Explanation:

F = ma

F = (936 kg)(9.8 m/s^2)

F = 9172.8 N

You might be interested in

A brick of mass 1.15 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle ha

stira [4]

Explanation:

It is given that,

Mass of the brick, m = 1.15 kg

Radius of the circle, r = 1.44 m

The cable will break if the tension exceeds 43.0 N

Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,

$T=\dfrac{mv^2}{r}+mg$

$\dfrac{T}{m}-g=\dfrac{v^2}{r}$

$v=\sqrt{(\dfrac{T}{m}-g)r}$

$v=\sqrt{(\dfrac{43}{1.15}-9.8)\times 1.44}$

v = 6.30 m/s

So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.

8 0

2 years ago

What is the desired tone of an instrument called?

Alik [6]

Answer:

Timbre also known as color or tone quantity which is used for music notes sound and tone are mostly know for guitars or pianos .

Does this help .

6 0

2 years ago

Read 2 more answers

What experimental evidence led to the development of this atomic model from the one before it?

Marina86 [1]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back

Explanation:

7 0

2 years ago

Read 2 more answers

The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the

Studentka2010 [4]

Answer:

$V(t)= 240V* H(t-5)$

Explanation:

The heaviside function is defined as:

$H(t) =1 \quad t\geq 0\\H(t) =0 \quad t$

so we see that the Heaviside function "switches on" when $t=0$ , and remains switched on when $t>0$

If we want our heaviside function to switch on when $t=5$ , we need the argument to the heaviside function to be 0 when $t=5$

Thus we define a function f:

$f(t) = H(t-5)$

The $-5$ term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)